Online Geometry theorems, problems, solutions, and related topics.
Proposed ProblemClick the figure below to see the complete problem 412 about Triangle, Angles, Congruence.See also:Complete Problem 412Level: High School, SAT Prep, College geometry
Another solution with full detail:http://img9.imageshack.us/img9/8220/problem412.pngFrom C draw CE//AB and CE=AB ( see attached picture)ABEC is a parallelogram and ∠ (ECy)= 2xTriangle DEC is isosceles => ∠ (DEC)=1/2∠ (ECy)=xQuadrilateral BDCE is cyclic with 2 opposite sides paralleledSo BDCE is an isosceles trapezoidAnd ∠ (BDA)= ∠ (ECy)=2x => Triangles ABD is isoscelesIn triangles ABD 4x=180-40 => x=35
My solution:Draw circle 0 (circumcircle of triangle ABD)Let E be the intersection of circle 0 with BC.Construct AE and DETriangle BED is isosceles since ∠EBD =∠EAD = x So ∠BAE = ∠BDE = xNow ∠BDC =40º+2x so ∠EDC=40º+x since ∠BDE= xTriangles ABE and EDC are similar by SASSo ∠ DCE= x , finally x=35º since x+x+40º+2x=180º => 4x=140º => x=35º
Let the angle bisector of A meet BC at E and let BE = pNow < DAE = < DBE = x so ABED is cyclic So (a-p)a = bc......(1)Angie bisector theorem p/(a-p) = c/b .......(2)(1) X (2) gives us ap = c^2 hence BC is tangential to Tr ACE at A and so < C = x Now we know all the angles of Tr. ABC in terms of x and adding 4x+ 40 = 180 and x = 35Sumith PeirisMoratuwaSri Lanka
Or far more simply because ABED is cyclic BE = DE and Tr.s ABE and CDE are congruent SAS and so < C = x and the result follows
Problem 412Brings BE=//DC then BECD is parallelogram so <BcE=x=< BAE=<BEA.Then BECA isisosceles trapezoid so <ACB=<AEB=x . Then 4x+40=180 so x=35.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE